Answer
The limit $\underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)$ is equal to $-1$.
Work Step by Step
Consider the provided limit,
$\underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)$
Simplify it as follows
$\begin{align}
& \underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( x\times 1-x\times \frac{1}{x} \right) \\
& =\underset{x\to 0}{\mathop{\lim }}\,\left( x-1 \right)
\end{align}$
Use limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)-g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)-\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$
$\begin{align}
& \underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( x-1 \right) \\
& =\underset{x\to 0}{\mathop{\lim }}\,x-\underset{x\to 0}{\mathop{\lim }}\,1
\end{align}$
Using limit property, $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c\text{ is constant}$
$\begin{align}
& \underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( x-1 \right) \\
& =\underset{x\to 0}{\mathop{\lim }}\,x-\underset{x\to 0}{\mathop{\lim }}\,1 \\
& =0-1 \\
& =-1
\end{align}$
Thus, $\underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)=-1$
